$f\left(x\right)=\frac{x^2}{x-4},x\ne 4$  

Wyznaczmy pochodną tej funkcji.   

$f{}^{\prime }\left(x\right)=\frac{\left(x^2\right){}^{\prime }\left(x-4\right)-x^2\left(x-4\right){}^{\prime }}{{\left(x-4\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{2x\left(x-4\right)-x^2\cdot 1}{{\left(x-4\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{2x^2-8x-x^2}{{\left(x-4\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{x^2-8x}{{\left(x-4\right)}^2}$  

 

Dla  $x=12$   otrzymujemy:

$f{}^{\prime }\left(12\right)=\frac{{12}^2-8\cdot 12}{{\left(12-4\right)}^2}$  

$f{}^{\prime }\left(12\right)=\frac{144-96}{8^2}$  

$f{}^{\prime }\left(12\right)=\frac{48}{64}$  

$f{}^{\prime }\left(12\right)=\frac{6}{8}$  

$f{}^{\prime }\left(12\right)=\frac{3}{4}$